3.5 Voltage Drop and Practical Calculation Setup

Voltage drop is about delivered voltage

Voltage drop is the reduction in voltage caused by conductor impedance as current flows. A load may be code-compliant for ampacity and overcurrent protection yet still perform poorly if the circuit is long and the voltage at the equipment is too low. Motors may run hot, lights may dim, and electronic equipment may misbehave.

The NEC includes informational guidance for reasonable voltage drop in branch circuits and feeders, but the exam may ask whether a given drop meets a design recommendation or may simply ask you to calculate the drop. Treat the specific question carefully. A voltage-drop recommendation is not the same as a mandatory ampacity rule unless a particular article, equipment instruction, or local amendment makes it enforceable.

Common formulas

One common single-phase formula using conductor resistance is:

VD = 2 x K x I x D / CM

One common three-phase formula is:

VD = 1.732 x K x I x D / CM

Where VD is voltage drop in volts, K is the conductor constant, I is current in amperes, D is one-way distance in feet, and CM is conductor area in circular mils. Common approximate K values are 12.9 for copper and 21.2 for aluminum, but always use the values supplied by the question or required reference if they differ.

Voltage drop percent is:

VD percent = (VD / circuit voltage) x 100

If you calculate a 6-volt drop on a 120-volt circuit, the drop is 6 / 120 x 100 = 5 percent. If the same 6-volt drop occurs on a 240-volt circuit, the percent is 2.5 percent. The volts dropped are the same, but the percentage is different.

Worked example: single-phase voltage drop

A 120-volt, single-phase, two-wire copper branch circuit supplies 16 A at a one-way distance of 75 ft. The conductor area is 6,530 circular mils. Use K = 12.9. Find approximate voltage drop.

VD = 2 x 12.9 x 16 x 75 / 6,530.

First multiply the numerator: 2 x 12.9 = 25.8. 25.8 x 16 = 412.8. 412.8 x 75 = 30,960. Divide by 6,530 = 4.74 V.

Percent drop = 4.74 / 120 x 100 = 3.95 percent.

The 2 in the formula accounts for the outgoing and return conductor path in a single-phase two-wire circuit. Do not double the 75 ft distance again when using this formula. If you use 150 ft for D and also keep the 2, the answer doubles incorrectly.

Worked example: three-phase voltage drop

A 480-volt, three-phase copper feeder carries 80 A for 150 ft one way. The conductor area is 41,740 circular mils. Use K = 12.9.

VD = 1.732 x 12.9 x 80 x 150 / 41,740.

1.732 x 12.9 = 22.3428. Multiply by 80 = 1,787.424. Multiply by 150 = 268,113.6. Divide by 41,740 = 6.42 V.

Percent drop = 6.42 / 480 x 100 = 1.34 percent.

A common wrong answer uses the single-phase 2 instead of 1.732. Another wrong answer divides by 277 because the system may be 277/480 V. If the feeder is serving a 480-volt three-phase load, use 480 as the circuit voltage for percent drop.

Conductor resistance method

Some questions give conductor resistance in ohms per 1,000 ft instead of circular mils. In that case, calculate resistance for the circuit length and use Ohm's Law.

For a single-phase two-wire circuit:

Total conductor resistance = resistance per 1,000 ft x (2 x one-way ft / 1,000)

VD = I x total resistance

Example: a conductor is 0.50 ohm per 1,000 ft. The one-way length is 200 ft and current is 20 A. Total conductor length for the current path is 400 ft. Resistance is 0.50 x 400 / 1,000 = 0.20 ohm. Voltage drop is 20 x 0.20 = 4 V.

This method makes the return path explicit, so do not also use the 2 x K formula. Use one method consistently.

Practical setup checklist

Before calculating, identify these facts:

1. Is the circuit single-phase or three-phase?
2. What is the circuit voltage at the load?
3. Is the conductor copper or aluminum?
4. What is the one-way length?
5. What current should be used: actual load current, calculated load, or design current?
6. What conductor size or circular-mil area is given?
7. Does the question ask for volts dropped, percent drop, or minimum conductor size?

If the question asks for minimum conductor size, you may need to rearrange the formula:

CM = 2 x K x I x D / VD for single-phase

CM = 1.732 x K x I x D / VD for three-phase

Then compare the required CM to conductor sizes. Choose a conductor with circular-mil area equal to or larger than the required value, subject to ampacity and installation rules. Voltage drop alone does not prove the conductor is code-compliant for ampacity.

Exam traps

The most common trap is length. Many formulas use one-way distance because the formula already includes the path multiplier. Read whether the question states one-way length or total conductor length. The second trap is material. Aluminum has higher resistance than copper, so using copper K for aluminum understates the voltage drop. The third trap is percent. A calculated 3.6 V drop is not 3.6 percent unless the circuit voltage is 100 V.

The fourth trap is treating informational recommendations as universal pass-fail rules. The ICC exam is based on listed references, including the applicable NEC edition for R17, T17, or G17. Use the code book for enforceable requirements and use voltage-drop calculations for the performance question being asked.

Calculator discipline

For a basic calculator, compute the numerator in chunks and write the result before dividing by circular mils. If the answer choices are close, carry two decimals. After calculating volts, always divide by system voltage and multiply by 100 if the answer asks for percent. If a percent answer is larger than 100 percent for an ordinary branch-circuit design problem, you likely entered circular mils or length incorrectly.